= Solution
Separate $\widetilde u=f(t)g(x)$. On a nonzero product region,
$$
\frac{f'}{\nu f}=\frac{g''}{g}=-\lambda.
$$
A decaying temporal factor requires $\lambda>0$, with $f=Ce^{-\nu\lambda t}$. The zero spatial boundary values give $g=A\sin(\sqrt\lambda x)$ and $\sin\sqrt\lambda=0$, hence $\lambda=j^2\pi^2$ with $j=1,2,\ldots$. Zero or negative separation parameters give no nonzero spatial mode with both boundary values zero.
The initial profile is already exactly the $j=n$ <sine> <eigenfunction>. Its coefficient therefore determines a single separated solution:
$$
\boxed{\widetilde u(x,t)=-\frac{G}{\nu n^2\pi^2}
e^{-\nu n^2\pi^2t}\sin(n\pi x)}.
$$
The exponential tends to zero because $\nu>0$. More generally <sine> <orthogonality> would determine a sum of such modes, but all other coefficients vanish here. This is the transient of the <heat equation with a single stationary sine forcing>.
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